We have water cooling chillers and crahs in our data center. What I want to learn is what will happen if I increase the chilled water temperature. For example, the indoor units (CRAH) set temperature is 23 degrees Celsius and the chiller's outlet set water temperature is 16 degrees Celsius. I want to set the chiller's outlet temperature to 21 Celsius to have more efficiency in terms of electrical usage. Are there any pros or cons to do this?

  • 1
    $\begingroup$ It will use less electricity but then it many not provide enough cooling. If there is a flow controller it may not use less electricity. $\endgroup$
    – paparazzo
    Sep 27, 2018 at 13:40
  • 1
    $\begingroup$ The down side would be the CRAH would have to be 3.5 times as big. Your 7 degree C delta is typical of air to fluid heat exchanger designs. I would consult with the system engineer who designed the system. I would experiment with changes of tenths of a degree and monitor carefully during adverse weather and load conditions. It may be beneficial to make small seasonal adjustments of a degree or so. You would need complete as-build schematics and parts lists to answer any better. Savings of more than a percent or two are unlikely. $\endgroup$
    – Phil Sweet
    Sep 28, 2018 at 0:12

2 Answers 2


Find the component cooled by the water that is closest to it's overheat temperature and that will tell you how much margin you have.

Do this test when under sustained load.

You could monitor the load such that under idle you set the water chiller to a higher temperature, and when you need high performance the chiller gets set to a lower temperature. (Of course this requires that you can monitor the load and can sett he temperature automatically)


View this as a heat exchanger at steady state. Your components generate $\dot{q}_g$ of heat flow. The water takes in that heat flow as $\dot{q}_w = \dot{m}_w \tilde{C}_{pw} (T_{wo} - T_{wi})$, with mass flow, specific heat, and temperatures defined. The remaining heat $\dot{q}_r = \dot{q}_g - \dot{q}_w$ heats the components.

You want to increase $T_{wi}$. One of three things can happen.

1) With constant mass flow of water, $T_{wo}$ will increase. This can happen when the water leaving the heat exchanger was initially cooler than the components that are to be cooled. In this case, you had some room to increase the inlet temperature because you have not yet over-run the outlet temperature of the water.

2) With constant mass flow of water, $T_{wo}$ will not change and $\dot{q}_w$ will decrease. This will happen when the water leaving the exchanger is already as hot as the components it is cooling. This is actually a case of a maximum heat exchange. In this case, when you increase $T_{wi}$, less heat is taken from the components. They will correspondingly get hotter as they reach a new equilibrium state. The output temperature of the water will then also get hotter, meaning less heat taken by the water. The result of this case will be that your components will get hotter, perhaps even hotter than you might have expected.

3) You will increase the mass flow, $\dot{m}_w$ to account for the increase in $T_{wo}$. The heat flow $\dot{q}_w$ will stay the same.

In all cases, you will decrease the electrical demand in the chiller.

Let's do a back of the envelope calculation. Assume the chiller runs at its thermodynamic efficiency. What is the difference in CoP for a refrigerator unit that must cool water at 23 oC to 16 oC versus one that must cool water at 23 oC to 21 oC.

$$ CoP = \frac{\dot{q}_C}{\dot{w}} = \frac{T_C}{T_H - T_C} $$

$$ CoP_o = \frac{289}{7} = 41 $$

$$ CoP_{new} = \frac{294}{2} = 147 $$

Assume the same mass flow of water (case 1 and case 2). The ratio of cooling needs is $\dot{q}_{new} / \dot{q}_o = \Delta T_{w,new} / \Delta T_{w,o} = 2/7$. The ratio of work loads is therefore found as

$$ \frac{\dot{w}_{new}}{\dot{w}_o} = \frac{\dot{q}_{new}}{\dot{q}_o} \frac{CoP_o}{CoP_{new}} = \frac{2}{7} \frac{41}{147} = 0.080$$

We find the power demand to a perfect chiller unit in the new case is predicted to be 8% of the power demand to a perfect chiller in the starting case. Inefficiencies in the chiller will increase this value.

In summary, you can decrease the electricity load at the chiller unit by running as per the first or second cases. Before you move forward, you should test the temperatures at the components that are being cooled relative to the output temperature of the cooling water. When the temperatures are the same, your savings will come at a cost that your components will get hotter.

The third case has to consider that $\dot{q}_{new} = \dot{q}_o$, so the maximum theoretical improvement will be 28%. The demand to the pump for the water will go up. Whether it offsets the gain is to be determined.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.